Electric Circuits, 9th Edition P30 pot

8,1 Introduction to the Natural Response of a Parallel RLC Circuit The first step in finding the natural response of the circuit shown in Fig.. The roots of the characteristic equation

c;

Figure 8

response

introducing these three forms, we show that the same forms apply

to the step response of a parallel RLC circuit as well as to the natu-ral and step responses of series RLC circuits

,1 A A circuit used to illustrate the natural

of a parallel RLC circuit

Figure 8

response

2 A A circuit used to illustrate the step

of a parallel RLC circuit

Figure 8.3 A A circuit used to illustrate the natural

response of a series RLC circuit

^ V

/ = 0

c

Figure 8.4 A A circuit used to illustrate the step

response of a series RLC circuit

8,1 Introduction to the Natural

Response of a Parallel RLC Circuit

The first step in finding the natural response of the circuit shown in Fig 8.1

is to derive the differential equation that the voltage v must satisfy We

choose to find the voltage first, because it is the same for each component After that, a branch current can be found by using the current-voltage relationship for the branch component We easily obtain the differential equation for the voltage by summing the currents away from the top node,

where each current is expressed as a function of the unknown voltage v:

(8.1)

We eliminate the integral in Eq 8.1 by differentiating once with respect to t,

and, because 70 is a constant, we get

1 dv v d v

(8.2)

We now divide through Eq 8.2 by the capacitance C and arrange the derivatives in descending order:

d v 1 dv v

~diI + lRClt+ Tc~ ' (8.3)

Comparing Eq 8.3 with the differential equations derived in Chapter 7 reveals that they differ by the presence of the term involving the second derivative Equation 8.3 is an ordinary, second-order differential equation with constant coefficients Circuits in this chapter contain both inductors and capacitors, so the differential equation describing these circuits is of the

sec-ond order Therefore, we sometimes call such circuits secsec-ond-order circuits

The General Solution of the Second-Order Differential Equation

We can't solve Eq 8.3 by separating the variables and integrating as we were able to do with the first-order equations in Chapter 7 The classical approach to solving Eq 8.3 is to assume that the solution is of exponential form, that is, to assume that the voltage is of the form

where A and s are unknown constants

Before showing how this assumption leads to the solution of Eq 8.3,

we need to show that it is rational The strongest argument we can make in favor of Eq 8.4 is to note from Eq 8.3 that the second derivative of the

solution, plus a constant times the first derivative, plus a constant times the

solution itself, must sum to zero for all values of t This can occur only if

higher order derivatives of the solution have the same form as the

solu-tion The exponential function satisfies this criterion A second argument

in favor of Eq 8.4 is that the solutions of all the first-order equations we

derived in Chapter 7 were exponential It seems reasonable to assume that

the solution of the second-order equation also involves the exponential

function

If Eq 8.4 is a solution of Eq 8.3, it must satisfy Eq 8.3 for all values of t

Substituting Eq 8.4 into Eq 8.3 generates the expression

, -, _, As v, Ae st

As 2 e st + e" + = 0,

RC LC

or

which can be satisfied for all values of t only if A is zero or the

parentheti-cal term is zero, because e xt ¥• 0 for any finite values of st We cannot use

A = 0 as a general solution because to do so implies that the voltage is

zero for all time —a physical impossibility if energy is stored in either the

inductor or capacitor Therefore, in order for Eq 8.4 to be a solution of

Eq 8.3, the parenthetical term in Eq 8.5 must be zero, or

s + —— + —— = 0 (8.6) A Characteristic equation, parallel

Equation 8.6 is called the characteristic equation of the differential

equa-tion because the roots of this quadratic equaequa-tion determine the

mathe-matical character of v(t)

The two roots of Eq 8.6 are

2RC V \ 2 R C / LC ; T ^ (8-7)

* = -55c-\/l^] - - ^ - (8-8)

If either root is substituted into Eq 8.4, the assumed solution satisfies the

given differential equation, that is, Eq 8.3 Note from Eq 8.5 that this

result holds regardless of the value of A Therefore, both

v = A^*1'and

v = Aif*

satisfy Eq 8.3 Denoting these two solutions v } and v 2 , respectively, we can

show that their sum also is a solution Specifically, if we let

v = Vi + v 2 = Â 1 ' + A 2 e Sl, (8.9)

then

dv

— = Aw**' + A 2 s 2 e s > 1 ,

- T = Arfe* + A 2 s 22 ệ

Substituting Eqs 8.9-8.11 into Eq 8.3 gives

1 ; l + ^ + Ic ) + A ^' H + i ^ + ^ l ' r

But each parenthetical term is zero because by definition st and s 2 are roots of the characteristic equation Hence the natural response of the

parallel RLC circuit shown in Fig 8.1 is of the form

v = A x e S]t + A 2 e $2t (8.13) Equation 8.13 is a repeat of the assumption made in Eq 8.9 We have

shown that v } is a solution, v 2 is a solution, and v x + v 2 is a solution Therefore, the general solution of Eq 8.3 has the form given in Eq 8.13

The roots of the characteristic equation (^ and s 2 ) are determined by the

circuit parameters /?, L, and C.The initial conditions determine the values

of the constants A] and A 2 Note that the form of Eq 8.13 must be

modi-fied if the two roots s\ and s 2 are equal We discuss this modification when

we turn to the critically damped voltage response in Section 8.2

The behavior of v(t) depends on the values of s-i and s 2 Therefore the

first step in finding the natural response is to determine the roots of the characteristic equation We return to Eqs 8.7 and 8.8 and rewrite them using a notation widely used in the literature:

-a + Va 2 — a>j),

.92 = —a — va2 wf),

(8.14)

(8.15)

where

Neper frequency, parallel RLC circuit •

a — 2RC 7

(8.16)

Resonant radian frequency, parallel

These results are summarized in Table 8.1

TABLE 8.1 Natural Response Parameters of the Parallel RLC Circuit

Parameter

«u

Terminology

Value In Natural Response Characteristic roots

Neper frequency Resonant radian frequency

S X = -a + Va2

s 2 = -a - Vcr

a = 2RC

1

"°~ Vie

- col

- (4

The exponent of e must be dimensionless, so both ,vj and s 2 (and

hence a and a>()) must have the dimension of the reciprocal of time, or

fre-quency To distinguish among the frequencies s h s 2 , a, and w(), we use the

following terminology: ,5¾ and s2 are referred to as complex frequencies, a

is called the neper frequency, and IOQ is the resonant radian frequency The

full significance of this terminology unfolds as we move through the

remaining chapters of this book All these frequencies have the

dimen-sion of angular frequency per time For complex frequencies, the neper

frequency, and the resonant radian frequency, we specify values using the

unit radians per second (rad/s) The nature of the roots s { and s 2 depends

on the values of a and o)() There are three possible outcomes First, if

<of) < a 2 , both roots will be real and distinct For reasons to be discussed

later, the voltage response is said to be overdamped in this case Second,

if col > a 2 , both s-\ and s 2 will be complex and, in addition, will be

conju-gates of each other In this situation, the voltage response is said to be

underdamped The third possible outcome is that co 2 ) = a 2 In this case, $i

and 52 will be real and equal Here the voltage response is said to be

critically damped As we shall see, damping affects the way the voltage

response reaches its final (or steady-state) value We discuss each case

separately in Section 8.2

Example 8.1 illustrates how the numerical values of S[ and s 2 are

determined by the values of R, L, and C

Example 8.1 Finding the Roots of the Characteristic Equation of a Parallel RLC Circuit

a) Find the roots of the characteristic equation that

governs the transient behavior of the voltage

shown in Fig 8.5 if R - 200 O, L = 50 mH, and

C = 0.2 ixF

b) Will the response be overdamped, underdamped,

or critically damped?

c) Repeat (a) and (b) for R = 312.5 H

d) What value of R causes the response to be

criti-cally damped?

Solution

a) For the given values of R, L, and C,

1 10f

a = 2RC (400)(0.2) = 1.25 X 104 rad/s,

2 l (io3)(io6) i n 8 2 ,

From Eqs 8.14 and 8.15,

s t = -1.25 X 104 + Vl.5625 X 108 - 108

= -12,500 + 7500 = -5000 rad/s,

Figure 8.5 A A circuit used to illustrate the natural response of

a parallel RLC circuit

fc = -1.25 x 104 - Vl.5625 X 108 - 108

b) The voltage response is overdamped because

o)() < a

c) Fori? = 312.511,

106

a = (625)(0.2) = 8000 rad/s

a 2 = 64 X 106 = 0.64 X 108rad2/s2

As col remains at 108 rad2/s2,

Si = -8000 + /6000 rad/s,

s 2 = -8000 - /6000 rad/s

(In electrical engineering, the imaginary number

V—T is represented by the letter /", because the

letter /' represents current.)

In this case, the voltage response is

under-damped since ag > a 2

d) For critical damping, a 2 = co 2h so

or

= 10\

and

R

i v

2RC)

1

2RC

106

1

LC '

= 104

(2 X 10 4 )(0.2) = 250 Q

I / A S S E S S M E N T PROBLEM

Objective 1—Be able to determine the natural response and the step response of parallel RLC circuits

8.1 The resistance and inductance of the circuit in

Fig 8.5 are 100 O and 20 mH, respectively

a) Find the value of C that makes the voltage

response critically damped

b) If C is adjusted to give a neper frequency of

5 krad/s, find the value of C and the roots of

the characteristic equation

c) If C is adjusted to give a resonant frequency

of 20 krad/s, find the value of C and the

roots of the characteristic equation

NOTE: Also try Chapter Problem 8.1

Answer: (a) 500 nF;

(b) C = 1 juF,

Si = -5000 + /5000 rad/s,

s 2 = -5000 - y'5000 rad/s;

(c) C = 125 nF, s?! = -5359 rad/s,

s 2 = -74,641 rad/s

8.2 The Forms of the Natural Response

of a Parallel RLC Circuit

So far we have seen that the behavior of a second-order RLC circuit depends

on the values of Vj and v2, which in turn depend on the circuit parameters R,

L, and C Therefore, the first step in finding the natural response is to calcu-late these values and, recalcu-latedly, determine whether the response is over-, under-, or critically damped

Completing the description of the natural response requires finding two

unknown coefficients, such as A { and A 2 in Eq 8.13.The method used to do this is based on matching the solution for the natural response to the initial conditions imposed by the circuit, which are the initial value of the current (or voltage) and the initial value of the first derivative of the current (or voltage) Note that these same initial conditions, plus the final value of the variable, will also be needed when finding the step response of a second-order circuit

In this section, we analyze the natural response form for each of the three types of damping, beginning with the overdamped response As we will see, the response equations, as well as the equations for evaluating the unknown coefficients, are slightly different for each of the three damping configurations This is why we want to determine at the outset of the problem whether the response is over-, under-, or critically damped

The Overdamped Voltage Response

When the roots of the characteristic equation are real and distinct, the

volt-age response of a parallel RLC circuit is said to be overdamped The

solu-tion for the voltage is of the form

A ^ + A 2 e Sl \ (8.18) < Voltage natural response—overdamped

parallel RLC circuit

where S] and s 2 are the roots of the characteristic equation The constants

A] and A 2 are determined by the initial conditions, specifically from the

values of v(Q + ) and dv(Q + )/dt, which in turn are determined from the

ini-tial voltage on the capacitor, V (h and the initial current in the inductor, /()

Next, we show how to use the initial voltage on the capacitor and the

initial current in the inductor to find A x and A 2 First we note from Eq 8.18

that A\ and A 2 First we note from Eq 8.18 that

v(0 + ) = A, + A 2 , (8.19) dv(Q + )

di = s x Ax + s 2 A 2 (8.20)

With S\ and s 2 known, the task of finding A { and A 2 reduces to finding

v(0 + ) and dv(0 + )/dt The value of v(0+) is the initial voltage on the

capac-itor V{\ We get the initial value of dv/dt by first finding the current in the

capacitor branch at t = 0+ Then,

dv(0 + ) i c (0 + )

We use Kirchhoff s current law to find the initial current in the

capac-itor branch We know that the sum of the three branch currents at t - 0+

must be zero The current in the resistive branch at t = ()+ is the initial

voltage Vo divided by the resistance, and the current in the inductive

branch is /() Using the reference system depicted in Fig 8.5, we obtain

After finding the numerical value of J*c(0+)i we use Eq 8.21 to find the

ini-tial value of dv/dt

We can summarize the process for finding the overdamped response,

v(t), as follows:

1 Find the roots of the characteristic equation, s } and s 2 , using the

val-ues of /?, L, and C

2 Find v(0 + ) and dv(0 + )/dt using circuit analysis

3 Find the values of Ai and A 2 by solving Eqs 8.23 and 8.24

simultaneously:

v(0 + ) = A t + A 2 , (8.23)

——— = ——- = s x A x + s 2 A 2 (8.24)

4 Substitute the values for s u s2, Aj, and A 2 into Eq 8.18 to

deter-mine the expression for v(t) for t > 0

Examples 8.2 and 8.3 illustrate how to find the overdamped response of a

parallel RLC circuit

Example 8.2 Finding the Overdamped Natural Response of a Parallel RLC Circuit

For the circuit in Fig 8.6, v(0 + ) = 12 V, and

'"L(0 + ) = 30 mA

a) Find the initial current in each branch of the

circuit

b) Find the initial value of dv/dt,

c) Find the expression for v(t)

d) Sketch v(t) in the interval 0 < r < 250 ms

Solution

a) The inductor prevents an instantaneous change

in its current, so the initial value of the inductor

current is 30 mA:

fe(0") = fe(0) = / L ( 0 + ) = 30mA

The capacitor holds the initial voltage across the

parallel elements to 12 V Thus the initial current

in the resistive branch, //?(0+), is 12/200, or

60 mA Kirchhoffs current law requires the sum

of the currents leaving the top node to equal

zero at every instant Hence

/c(o+) = -/L(o+) - ;*(o+)

= - 9 0 mA

Note that if we assumed the inductor current and

capacitor voltage had reached their dc values at

the instant that energy begins to be released,

*c(0_) = 0 In other words, there is an

instanta-neous change in the capacitor current alt = 0

b) Because ic — C(dv/dt),

dv(() + ) - 9 0 X 10"

dt 0.2 X 10- 6

- 4 5 0 kV/s

c) The roots of the characteristic equation come

from the values of R, L, and C For the values

specified and from Eqs 8.14 and 8.15 along with

8.16 and 8.17,

-1.25 X 104 + Vl.5625 X 108 - 108

-12,500 + 7500 = -5000rad/s,

s 2 = -1.25 X 104 - 2 1.5625 X 108 - 108

-12,500 - 7500 = -20,000 rad/s

Figure 8.6 A The circuit for Example 8.2

Because the roots are real and distinct, we know that the response is overdamped and hence has

the form of Eq 8.18 We find the co-efficients A x

and A 2 from Eqs 8.23 and 8.24 We've already

determined s\, s 2 , v(0 + ), and dv(Q + )/dt, so

12 = A l + A 2 ,

- 4 5 0 X 103 = - 5 0 ( ) 0 ^ - 20,000A2

We solve two equations for A r and A 2 to obtain

Ai = - 1 4 V and A 2 = 26 V Substituting these values into Eq 8.18 yields the overdamped volt-age response:

v(t) = (-Ue' 5i)m + 26e-20000') V, t > 0

As a check on these calculations, we note that

the solution yields v(0) = 12 V and dv(0 + )/dt

= -450,000 V/s

d) Figure 8.7 shows a plot of v(t) versus t over the interval 0 < t < 250 ms

t(/XS)

Figure 8.7 • The voltage response for Example 8.2

Calculating Branch Currents in the Natural Response of a Parallel RLC Circuit

Derive the expressions that describe the three

branch currents i R , /L, and ic in Example 8.2

(Fig 8.6) during the time the stored energy is being

released

Solution

We know the voltage across the three branches

from the solution in Example 8.2, namely

,-500()/ -20,000/

v(t) = (-14*?-™"" + 26<r^u u u') V, t > 0

The current in the resistive branch is then

i R (t) = ^ - (-70e- 5m)l + 130*-20'000') mA, t > 0

There are two ways to find the current in the

induc-tive branch One way is to use the integral

relation-ship that exists between the current and the voltage

at the terminals of an inductor:

hiO = 7 / v L (x)dx + /()

A second approach is to find the current in the capacitive branch first and then use the fact that

ia + ijL + 'c = 0 Let's use this approach The cur-rent in the capacitive branch is

ic( 0 = c dv

dt

0.2 X l(r6(70,000e" -5000/ 20,000/

520,000<r z,WJUW )

(14c -5000/ 104^-20.000/)mA^ , >( )+

Note that *c(0+) = - 9 0 mA, which agrees with the result in Example 8.2

Now we obtain the inductive branch current from the relationship

k(0 = -<K(0 - W0

(56c -5000/ 2 ^ - 2 0 0 0 0 / ) m A ? / s 0

We leave it to you, in Assessment Problem 8.2, to show that the integral relation alluded to leads to

the same result Note that the expression for i L

agrees with the initial inductor current, as it must

^ A S S E S S M E N T P R O B L E M S

Objective 1—Be able to determine the natural response and the step response of parallel RLC circuits

8.2 Use the integral relationship between i L and v

to find the expression for i L in Fig 8.6

,-5000/ 20,000/N

Answer: i L (t) = ( 5 6 ^ ^ - 2de~ mwx ) mA, t > 0

8.3 The element values in the circuit shown are

R = 2 kH, L = 250 mH, and C = 10 nF The

initial current /0 in the inductor is —4 A, and

the initial voltage on the capacitor is 0 V The

output signal is the voltage v Find (a) //?(0+);

(b) /c(0+); (c) dv(0 + )/dt; (d) AX; (e) A 2 ; and

(f) v{t) when t > 0

Answer: (a) 0;

(b)4A;

(c) 4 X 108 V/s;

(d) 13,333 V;

(e) -13,333 V;

(f) 13,333(e-iao00' — e -40.000/ ) V

NOTE: Also try Chapter Problems 8.8, 8.11, and 8.18

The Underdamped Voltage Response

When IOQ > a2, the roots of the characteristic equation are complex, and the response is underdamped For convenience, we express the roots jj

and Si as

(8.25) (8.26)

where

The term (o d is called the damped radian frequency We explain later the

reason for this terminology

The underdamped voltage response of a parallel RLC circuit is

Voltage natural response—underdamped

parallel RLC circuits • v(t) = B\e~ at cos <o d t + B 2 e~ at sin w d t, (8.28)

which follows from Eq 8.18 In making the transition from Eq 8.18 to

Eq 8.28, we use the Euler identity:

Thus,

v{t) = A^- n+l "' l)l + A 2 e~ {a+Mt

= Aie-^ei"* + A 2 e~ at e~^

= e~"'(Ai cos co d t + / A i sin o>/ + A 2 coso) (t t - jA 2 sin (o d t)

= e~°"[(Ai + A 2 ) cosco d t + j{A { - A 2 )smcD d t]

At this point in the transition from Eq 8.18 to 8.28, replace the arbitrary

constants A\ + A 2 and }(A\ - A 2 ) with new arbitrary constants denoted

B x and B 2 to get

v = e~ m (B] cosco d t + B 2 sma) d t)

= B x e~°" cos co d t + B 2 e~ at sin a) d t

The constants B x and B 2 are real, not complex, because the voltage is a

real function Don't be misled by the fact that B 2 = j(Ai - A 2 ) In this

underdamped case, A j and A 2 are complex conjugates, and thus B { and B 2

are real (See Problems 8.12 and 8.13.) The reason for defining the

under-damped response in terms of the coefficients B x and B 2 is that it yields a

sim-pier expression for the voltage, v We determine B { and B 2 by the initial

energy stored in the circuit, in the same way that we found A x and A 2 for the

overdamped response: by evaluating v at t = 0+ and its derivative at t = 0+

As with S\ and y2, a and oj d are fixed by the circuit parameters R, L, and C

For the underdamped response, the two simultaneous equations that

determine B { and B 2 are

v(0 + ) = V Q = 0, (8.30)

dv(0 + ) i c (0 + )

Let's look at the general nature of the underdamped response First,

the trigonometric functions indicate that this response is oscillatory; that

is, the voltage alternates between positive and negative values The rate at

which the voltage oscillates is fixed by a) d Second, the amplitude of the

oscillation decreases exponentially The rate at which the amplitude falls

off is determined by a Because a determines how quickly the oscillations

subside, it is also referred to as the damping factor or damping coefficient

That explains why a> d is called the damped radian frequency If there is no

damping, a = 0 and the frequency of oscillation is a) {) Whenever there is a

dissipative element, R, in the circuit, a is not zero and the frequency of

oscillation, a> d , is less than o>0 Thus when a is not zero, the frequency of

oscillation is said to be damped

The oscillatory behavior is possible because of the two types of

energy-storage elements in the circuit: the inductor and the capacitor (A

mechan-ical analogy of this electric circuit is that of a mass suspended on a spring,

where oscillation is possible because energy can be stored in both the

spring and the moving mass.) We say more about the characteristics of the

underdamped response following Example 8.4, which examines a circuit

whose response is underdamped In summary, note that the overall

process for finding the underdamped response is the same as that for the

overdamped response, although the response equations and the

simulta-neous equations used to find the constants are slightly different

Example 8.4 Finding the Underdamped Natural Response of a Parallel RLC Circuit

In the circuit shown in Fig 8.8, V {) = 0, and

/() = -12.25 mA

a) Calculate the roots of the characteristic equation

b) Calculate v and dv/dt at t = 0+

c) Calculate the voltage response for t S: 0

d) Plot v{t) versus t for the time interval

( ) < / < ! ! ms

0.125 /xF

Figure 8.8 • The circuit for Example 8.4

Solution

a) Because

2RC 2(20) 103(0.125)

0)() =

we have

10f (8)(0.125)

200 rad/s,

103 rad/s

a>5 > a~